Problem 29 - A First Course in Differential Equations with Modeling Applications

Question

Solve the given initial-value problem. Use a graphing utility to graph the solution curve. $$x y^{\prime \prime}+y^{\prime}=x, \quad y(1)=1, y^{\prime}(1)=-\frac{1}{2}$$

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{x^2}{4} - \ln|x| + \frac{3}{4} \).

Step 1: Understand the Differential Equation

The given differential equation is a second-order linear non-homogeneous equation: \( x y'' + y' = x \). The problem provides initial conditions, making it an initial-value problem (IVP).

Step 2: Simplify the Equation

Divide the entire equation by \( x \) to simplify it: \[ y'' + \frac{y'}{x} = 1 \]. This form helps in identifying suitable methods for solving the equation.

Step 3: Use an Integrating Factor

Recognize this as a first-order linear equation in \( y' \): \[ y'' + \frac{y'}{x} = 1 \]. To solve it, treat \( y' \) as a function and multiply by an integrating factor, \( \mu(x) = e^{\int \frac{1}{x} \, dx} = x \), to make the left side an exact derivative.

Step 4: Transform and Integrate

Multiply through by \( x \): \[ x y'' + y' = x \]. It becomes \( x y'' + \frac{x}{x}y' = x \), or \( (xy')' = x \). Integrate both sides with respect to \( x \): \[ xy' = \frac{x^2}{2} + C \], where \( C \) is a constant.

Step 5: Solve for \( y' \)

Solve for \( y' \): \[ y' = \frac{x}{2} + \frac{C}{x} \]. Use the initial condition \( y'(1) = -\frac{1}{2} \) to find \( C \): \(-\frac{1}{2} = \frac{1}{2} + \frac{C}{1}\). This gives \( C = -1 \).

Step 6: Integrate to Find \( y \)

Substitute \( C = -1 \) into the equation for \( y' \) and integrate: \[ y' = \frac{x}{2} - \frac{1}{x} \]. Then, \( y = \int \left( \frac{x}{2} - \frac{1}{x} \right) \, dx \).

Step 7: Evaluate the Integral

Calculate the integral: \[ y = \frac{x^2}{4} - \ln|x| + D \], where \( D \) is a constant of integration. Apply the initial condition \( y(1) = 1 \) to solve for \( D \).

Step 8: Apply Initial Condition to Solve for Constant

Substitute \( x=1 \), \( y(1)=1 \) into the equation \( y(1) = \frac{1^2}{4} - \ln(1) + D \) to find \( D \). Since \( \ln(1) = 0 \), \( 1 = \frac{1}{4} + D \) thus, \( D = \frac{3}{4} \).

Step 9: Final Solution Function

The function \( y \) that satisfies both the differential equation and the initial conditions is \[ y(x) = \frac{x^2}{4} - \ln|x| + \frac{3}{4} \]. Ensure to verify this solution by differentiating and checking with initial conditions.

Step 10: Graph the Solution Curve

Using a graphing utility, plot \( y(x) = \frac{x^2}{4} - \ln|x| + \frac{3}{4} \) to visualize the solution curve. You should see the curve passing through point \((1, 1)\) and indicating the slope at that point as \(-\frac{1}{2}\).

Key Concepts

Initial Value Problem

An initial value problem (IVP) is a specific type of differential equation. It involves finding a function that satisfies a differential equation and also meets certain initial conditions. These conditions are defined at a specific point, typically written as \(y(x_0) = y_0\) and \(y'(x_0) = y'_0\). For the problem at hand, we need to solve the differential equation \(x y'' + y' = x\) while satisfying the conditions \(y(1)=1\) and \(y'(1)=-\frac{1}{2}\).
The initial conditions help ensure that out of the infinite number of possible solutions for a differential equation, we find the one unique solution that fits the specific problem characteristics.

Second-Order Linear Equation

Second-order linear differential equations are characterized by the highest derivative being the second derivative. They are often written in the standard form \(a(x)y'' + b(x)y' + c(x)y = f(x)\). Here, our equation is \(x y'' + y' = x\).
Such equations are important in various fields like physics and engineering, as they often model systems with memory or inertia (such as systems governed by Newton's second law of motion). Converting the equation into a simpler form, as done by dividing through by \(x\), can make it easier to solve using specific techniques, such as applying an integrating factor.

Integrating Factor

An integrating factor is a mathematical tool used to solve differential equations, particularly linear first-order equations, by transforming them into exact equations. Here, the equation \(y'' + \frac{y'}{x} = 1\) is recognized as a first-order linear equation in \(y'\).
The integrating factor method involves multiplying the entire equation by a function \(\mu(x)\), which is derived from the equation\( \mu(x) = e^{\int \frac{1}{x} \, dx} = x\). This strategy transforms the left-hand side of the equation into a derivative of a product, making it easier to integrate and find the solution.

Solution Curve Visualization

Solution curve visualization involves graphing the solution of a differential equation to better understand its behavior. In this problem, the solution \(y(x) = \frac{x^2}{4} - \ln|x| + \frac{3}{4}\) is graphed.
This visual representation is crucial for interpreting the solution beyond just the algebraic expression. It allows us to see key aspects like asymptotic behavior, where the curve is heading or where it is flat, and how initial conditions like \((1, 1)\) and the slope \(-\frac{1}{2}\) at that point apply to the overall behavior of the function.

• Graphing helps verify the solution visually.
• It provides insight into the real-world interpretation of the mathematical model.